Question: $\text E = \left[\begin{array}{r}5 \\ 2 \\ -2\end{array}\right]$ and $\text D = \left[\begin{array}{rr}5 & 3\end{array}\right]$ Let $\text {H = ED}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{5} \\ 2 \\ -2\end{array}\right]\left[\begin{array}{rr} {5} & 3\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(5)\cdot(5)\\\\ &=25 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A ${H}_{2,1}$ does not exist. (Choice B) B $5 \cdot 3 = 15$ (Choice C) C $2 \cdot 5= 10$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}25 & 15 \\ 10 & 6 \\ -10 & -6\end{array}\right] $